Question

Solve the expression $\sin \left( 6\theta \right)+\sin \left( 2\theta \right)=0$ for $\theta $.

Answer 1

Hint: We will use the formula of trigonometry here. These formulas are given by $\sin A+\cos B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$, $\sin \left( \theta \right)=0$ which results into $\theta =n\pi $for all n belongs to integers and $\cos \left( \theta \right)=0$ which results into $\theta =2n\pi \pm \dfrac{\pi }{2}$ where n belongs to integers. This is because with the help of these formulas we will be able to find the value of $\theta $.

Complete step-by-step answer:
Now, we will consider the trigonometric equation $\sin \left( 6\theta \right)+\sin \left( 2\theta \right)=0...(i)$. We will consider the left hand side of the equation (i). Therefore, we now have $\sin \left( 6\theta \right)+\sin \left( 2\theta \right)$. Now, we will use the formula given by $\sin A+\cos B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$. Therefore, we get $\begin{align}
  & \sin \left( 6\theta \right)+\sin \left( 2\theta \right)=2\sin \left( \dfrac{6\theta +2\theta }{2} \right)\cos \left( \dfrac{6\theta -2\theta }{2} \right) \\
 & \Rightarrow \sin \left( 6\theta \right)+\sin \left( 2\theta \right)=2\sin \left( \dfrac{8\theta }{2} \right)\cos \left( \dfrac{4\theta }{2} \right) \\
 & \Rightarrow \sin \left( 6\theta \right)+\sin \left( 2\theta \right)=2\sin \left( 4\theta \right)\cos \left( 2\theta \right) \\
\end{align}$
Now since we have $\sin \left( 6\theta \right)+\sin \left( 2\theta \right)=0$ therefore we will also have $2\sin \left( 4\theta \right)\cos \left( 2\theta \right)=0$. Now these will work as factors. Therefore, we now have $\sin \left( 4\theta \right)=0$ and $\cos \left( 2\theta \right)=0$. So, now we will use the formula $\sin \left( \theta \right)=0$ which results into $\theta =n\pi $for all n belongs to integers. Therefore we have $4\theta =n\pi $ or $\theta =\dfrac{n\pi }{4}$which belongs to all integers. Now we will substitute the value of n = 1 in the equation $\theta =\dfrac{n\pi }{4}$ which gives us the value as $\theta =\dfrac{\pi }{4}$.
Now to simplify $\cos \left( 2\theta \right)=0$ we will apply the formula given by $\cos \left( \theta \right)=0$ which results into $\theta =2n\pi \pm \dfrac{\pi }{2}$ where n belongs to integers. Therefore, we have $\cos \left( 2\theta \right)=0$ as $2\theta =2n\pi \pm \dfrac{\pi }{2}$ where n belong to the integers. This can be further written as $\theta =n\pi \pm \dfrac{\pi }{4}$ where n = ...,- 1, 1, 0, - 2, 2... .
Now we will substitute the value of n = 1 in the equation $\theta =n\pi \pm \dfrac{\pi }{4}$ which gives us the value as $\theta =\pi \pm \dfrac{\pi }{4}$. As here we are not given the quadrants as in which quadrant the $\theta $ lies. Therefore, we will consider both the values of it. Thus we have $\theta =\pi +\dfrac{\pi }{4}$ and $\theta =\pi -\dfrac{\pi }{4}$. After simplification we get $\theta =\dfrac{5\pi }{4}$ and $\theta =\dfrac{3\pi }{4}$.
Hence, the values of $\theta $ for the trigonometric expression $\sin \left( 6\theta \right)+\sin \left( 2\theta \right)=0$ are $\theta =\dfrac{\pi }{4},~\theta =\dfrac{5\pi }{4}$ and $\theta =\dfrac{3\pi }{4}$.

Note: We have not taken here the generalized solutions for the trigonometric expression $\sin \left( 6\theta \right)+\sin \left( 2\theta \right)=0$. This is because it is not given in the question as specified. So, this is why we have substituted the value of n = 1 in this question. The use of the formula $\sin \left( 2\theta \right)=2\sin \theta \cos \theta $ will be of no use here. As we can see that although, the formula fits in the given trigonometric expression. But it will not lead to the solution. The value of the angle $\theta =n\pi \pm \dfrac{\pi }{4}$ is separated into $\theta =\pi +\dfrac{\pi }{4}$ and $\theta =\pi -\dfrac{\pi }{4}$. This is due to the fact that $\cos \left( \dfrac{\pi }{2} \right)=0$. And the angle $\dfrac{\pi }{2}$ lies in the second as well as the third quadrant. Thus, we have such formula $\theta =n\pi \pm \dfrac{\pi }{4}$ for $\cos \left( 2\theta \right)=0$.